Let a be the side length of the regular hexagon.
Calculating a.
(14/sin60) = (16/sinb)
b = 81.7867892983°
c = 180-60-b
c = 38.2132107017°
It implies;
(a/sin38.2132107017) = (14/sin...
Observing similar plane shape (scalene triangle) side length ratios.
Therefore;
y - 8
20 - 22
It implies;
(y/20) = (8/22)
(y/20) = (4/11)
Cross Multiply.
11y = 80
y = 80/11 units.
y = 7.2...
3² = 2c²-2c²cos108
108° is the single interior angle of the regular pentagon.
It implies;
9 = 2.6180339887c²
c² = 3.4376941013
c = 1.8541019663 units.
c is the side length of the regular pent...
Let the side length of the regular pentagon be 1 unit.
a² = 2-2cos108
108° is the single interior angle of the regular pentagon.
a = 1.6180339887 units.
sin72 = b/1
b = 0.9510565163 units.
cos72...
Let the side length of the regular pentagon be 1 unit.
a² = 2-2cos108
108° is the single interior angle of the regular pentagon.
a = 1.6180339887 units.
Area Blue is;
0.5*1*1.6180339887s...
It implies;
2a² = √(2)²
Where a is the radius of the half circle.
Therefore;
2a² = 2
a² = 1
a = 1 unit.
Again, a is the radius of the half circle.
x (diameter of the half circle) is;...
b = (180(12-2))/12
b = 1800/12
b = 150°
b is the single interior angle of the regular dodecagon.
2c+2(150) = 360
2c = 60
c = 30°
d = 150-15-30
d = 105°
2e+3(150) = 180(5-2)
2e = 540-450
2e = 90
e...
Let the inscribed square side be 1 unit.
sin30 = a/1
a = ½ units.
a = 0.5 units.
a is the side length of one of the four congruent isosceles inscribed shaded triangle.
sin60 = b/1
b = ½√(3) units....
Let a be half AB.
b = ½(3+7+2)
b = 6 units.
It implies;
6² = a²+c²
c² = 36-a²
c = √(36-a²) units.
d = ½(7)
d = 3.5 units.
e = 4-3.5
e = 0.5 units.
Therefore;
c² = 0.5²+f²...
a² = 12²+5²
a = √(144+25)
a = 13 cm.
a is AD.
12b+5b+13b = 12*5
30b = 60
b = 2 cm.
b is the radius of the inscribed circle.
tanc = 12/5
c = atan(12/5)°
tan(0.5c) = 2/d
⅔ = 2/d
d =...
